[00:15] Here, I want to discuss one common type of problem where integration comes up,
[00:19] finding the average of a continuous variable.
[00:23] This is a perfectly useful thing to know in its own right,
[00:26] but what's really neat is that it can give us a completely different
[00:29] perspective for why integrals and derivatives are inverses of each other.
[00:33] To start, take a look at the graph of sinx between 0 and pi, which is half of its period.
[00:40] What is the average height of this graph on that interval?
[00:44] It's not a useless question.
[00:46] All sorts of cyclic phenomena in the world are modeled using sine waves.
[00:50] For example, the number of hours the sun is up per day as a
[00:54] function of what day of the year it is follows a sine wave pattern.
[00:58] So if you wanted to predict the average effectiveness of solar panels in summer months vs.
[01:04] winter months, you'd want to be able to answer a question like this,
[01:08] what is the average value of that sine function over half of its period?
[01:13] Where as a case like this is going to have all sorts of constants mucking up the
[01:18] function, you and I are going to focus on a pure, unencumbered sinx function,
[01:22] but the substance of the approach would be totally the same in any other application.
[01:28] It's kind of a weird question to think about though, isn't it?
[01:31] The average of a continuous variable.
[01:33] Usually with averages we think of a finite number of variables,
[01:37] where you can add them all up and divide that sum by how many there are.
[01:44] But there are infinitely many values of sinx between 0 and pi,
[01:48] and it's not like we can just add up all those numbers and divide by infinity.
[01:54] This sensation comes up a lot in math, and it's worth remembering,
[01:58] where you have this vague sense that you want to add together infinitely
[02:02] many values associated with a continuum, even though that doesn't make sense.
[02:08] And almost always, when you get that sense, the key is to use an integral somehow.
[02:13] And to think through exactly how, a good first step is to
[02:17] just approximate your situation with some kind of finite sum.
[02:20] In this case, imagine sampling a finite number of points evenly spaced along this range.
[02:27] Since it's a finite sample, you can find the average by just adding up all the heights
[02:32] sinx at each one of these, and then dividing that sum by the number of points you sampled.
[02:39] And presumably, if the idea of an average height among all infinitely many
[02:43] points is going to make any sense at all, the more points we sample,
[02:47] which would involve adding up more and more heights,
[02:50] the closer the average of that sample should be to the actual average of
[02:54] the continuous variable.
[02:57] And this should feel at least somewhat related to taking an integral of sinx
[03:01] between 0 and pi, even if it might not be exactly clear how the two ideas match up.
[03:07] For that integral, remember, you also think of a sample of inputs on this continuum,
[03:13] but instead of adding the height sinx at each one and dividing by how many there are,
[03:18] you add up sinx times dx, where dx is the spacing between the samples.
[03:24] That is, you're adding up little areas, not heights.
[03:28] And technically, the integral is not quite this sum,
[03:31] it's whatever that sum approaches as dx approaches 0.
[03:35] But it is actually quite helpful to reason with respect to one of these finite
[03:39] iterations, where we're looking at a concrete size for dx and some specific number of
[03:44] rectangles.
[03:45] So what you want to do here is reframe this expression for the average,
[03:50] this sum of the heights divided by the number of sampled points,
[03:54] in terms of dx, the spacing between samples.
[03:59] And now, if I tell you that the spacing between these points is, say, 0.1,
[04:04] and you know that they range from 0 to pi, can you tell me how many there are?
[04:11] Well, you can take the length of that interval, pi,
[04:14] and divide it by the length of the space between each sample.
[04:19] If it doesn't go in perfectly evenly, you'd have to round down to the nearest integer,
[04:23] but as an approximation, this is completely fine.
[04:27] So if we write that spacing between samples as dx,
[04:31] the number of samples is pi divided by dx.
[04:34] And when we substitute that into our expression up here,
[04:38] you can rearrange it, putting that dx up top and distributing it into the sum.
[04:43] But think about what it means to distribute that dx up top.
[04:48] It means that the terms you're adding up will look like
[04:51] sinx times dx for the various inputs x that you're sampling.
[04:56] So that numerator looks exactly like an integral expression.
[04:59] And so for larger and larger samples of points,
[05:02] this average will approach the actual integral of sinx between 0 and pi,
[05:07] all divided by the length of that interval, pi.
[05:11] In other words, the average height of this graph is this area divided by its width.
[05:18] On an intuitive level, and just thinking in terms of units,
[05:21] that feels pretty reasonable, doesn't it?
[05:23] Area divided by width gives you an average height.
[05:26] So with this expression in hand, let's actually solve it.
[05:31] As we saw last video, to compute an integral, you need to find an antiderivative
[05:36] of the function inside the integral, some other function whose derivative is sinx.
[05:42] And if you're comfortable with derivatives of trig functions,
[05:45] you know that the derivative of cosine is negative sine.
[05:49] So if you just negate that, negative cosine is the function we want,
[05:53] the antiderivative of sine.
[05:55] And to gut-check yourself on that, look at this graph of negative cosine.
[06:00] At 0, the slope is 0, and then it increases up to some maximum slope at pi halves,
[06:06] and then goes back down to 0 at pi.
[06:09] And in general, its slope does indeed seem to
[06:12] match the height of the sine graph at every point.
[06:17] So what do we have to do to evaluate the integral of sine between 0 and pi?
[06:22] We evaluate this antiderivative at the upper bound,
[06:25] and subtract off its value at the lower bound.
[06:29] More visually, that's the difference in the height of
[06:32] this negative cosine graph above pi and its height at 0.
[06:37] And as you can see, that change in height is exactly 2.
[06:41] That's kind of interesting, isn't it?
[06:43] That the area under this sine graph turns out to be exactly 2?
[06:48] So the answer to our average height problem, this integral divided by the width
[06:53] of the region, evidently turns out to be 2 divided by pi, which is around 0.64.
[07:01] I promised at the start that this question of finding the average of a function offers
[07:06] an alternate perspective on why integrals and derivatives are inverses of each other,
[07:11] why the area under one graph has anything to do with the slope of another graph.
[07:16] Notice how finding this average value, 2 divided by pi,
[07:20] came down to looking at the change in the antiderivative,
[07:24] negative cosine x, over the input range, divided by the length of that range.
[07:30] And another way to think about that fraction is as the rise over run slope between
[07:35] the point of the antiderivative graph below 0 and the point of that graph above pi.
[07:41] Think about why it might make sense that this slope would
[07:45] represent an average value of sine of x on that region.
[07:50] By definition, sine of x is the derivative of this antiderivative graph,
[07:55] giving us the slope of negative cosine at every point.
[07:59] Another way to think about the average value of sine of x is
[08:03] as the average slope over all tangent lines between 0 and pi.
[08:08] And when you view things like that, doesn't it make a lot of sense
[08:12] that the average slope of a graph over all its points in a certain
[08:16] range should equal the total slope between the start and end points?
[08:23] To digest this idea, it helps to think about what it looks like for a general function.
[08:28] For any function f of x, if you want to find its average value on some interval,
[08:33] say between a and b, what you do is take the integral of f on that
[08:38] interval divided by the width of that interval, b minus a.
[08:43] You can think of this as the area under the graph divided by its width,
[08:47] or more accurately, it is the signed area of that graph,
[08:50] since any area below the x-axis is counted as negative.
[08:55] And it's worth taking a moment to remember what this area has to do with the usual notion
[09:00] of a finite average, where you add up many numbers and divide by how many there are.
[09:05] When you take some sample of points spaced out by dx,
[09:08] the number of samples is about equal to the length of the interval divided by dx.
[09:14] So if you add up the values of f of x at each sample and divide by
[09:18] the total number of samples, it's the same as adding up the product
[09:23] f of x times dx and dividing by the width of the entire interval.
[09:27] The only difference between that and the integral is that the integral asks
[09:32] what happens as dx approaches 0, but that just corresponds with samples of
[09:36] more and more points that approximate the true average increasingly well.
[09:42] Now for any integral, evaluating it comes down to finding an antiderivative of f of x,
[09:48] commonly denoted capital F of x.
[09:51] What we want is the change to this antiderivative between a and b,
[09:56] capital F of b minus capital F of a, which you can think of as
[10:00] the change in height of this new graph between the two bounds.
[10:06] I've conveniently chosen an antiderivative that passes through 0 at the lower bound here,
[10:11] but keep in mind you can freely shift this up and down adding whatever
[10:16] constant you want and it would still be a valid antiderivative.
[10:21] So the solution to the average problem is the change in the height of
[10:25] this new graph divided by the change to the x value between a and b.
[10:31] In other words, it is the slope of the antiderivative graph between the two endpoints.
[10:37] And again, when you stop to think about it, that should make a lot of sense,
[10:41] because little gives us the slope of the tangent line to this graph at each point.
[10:47] After all, it is by definition the derivative of capital F.
[10:52] So why are antiderivatives the key to solving integrals?
[10:57] My favorite intuition is still the one I showed last video,
[11:01] but a second perspective is that when you reframe the question of finding an average of
[11:06] a continuous value as instead finding the average slope of a bunch of tangent lines,
[11:11] it lets you see the answer just by comparing endpoints,
[11:15] rather than having to actually tally up all the points in between.
[11:23] In the last video I described a sensation that should bring integrals to your mind,
[11:27] namely if you feel like the problem you're solving could be approximated by
[11:31] breaking it up somehow and adding up a large number of small things.
[11:36] Here I want you to come away recognizing a second
[11:38] sensation that should also bring integrals to your mind.
[11:42] If ever there's some idea that you understand in a finite context,
[11:46] and which involves adding up multiple values, like taking the average of a
[11:51] bunch of numbers, and if you want to generalize that idea to apply to an infinite
[11:56] continuous range of values, try seeing if you can phrase things in terms of an integral.
[12:02] It's a feeling that comes up all the time, especially in probability,
[12:05] and it's definitely worth remembering.
[12:09] My thanks, as always, go to those making these videos possible.
[12:31] Thank you.